# Finding the 9th derivative of $\frac{\cos(5 x^2)-1}{x^3}$

How do you find the 9th derivative of $(\cos(5 x^2)-1)/x^3$ and evaluate at $x=0$ without differentiating it straightforwardly with the quotient rule? The teacher's hint is to use Maclaurin Series, but I don't see how.

My attempts at deriving yielded this:

$$-10\sin(5x^2)/x^2 - 3(\cos(5x^2) - 1)/x^4$$

$$-100\cos(5x^2)/x + 50\sin(5x^2)/x^3 + 12(\cos(5x^2) - 1)/x^5$$

$$1000\sin(5x^2) + 600\cos(5x^2)/x^2 - 270\sin(5x^2)/x^4 - 60(\cos(5x^2) - 1)/x^6$$

As a programmer, I used sympy to calculate the derivative to be

$$- 1000000000 x^{6} \sin{\left (5 x^{2} \right )} + 900000000 x^{4} \cos{\left (5 x^{2} \right )} + 540000000 x^{2} \sin{\left (5 x^{2} \right )} + 378000000 \cos{\left (5 x^{2} \right )} - 472500000 \frac{\sin{\left (5 x^{2} \right )}}{x^{2}} - 481950000 \frac{\cos{\left (5 x^{2} \right )}}{x^{4}} + 393120000 \frac{\sin{\left (5 x^{2} \right )}}{x^{6}} + 240408000 \frac{\cos{\left (5 x^{2} \right )}}{x^{8}} - 97977600 \frac{\sin{\left (5 x^{2} \right )}}{x^{10}} - 19958400 \frac{\cos{\left (5 x^{2} \right )} - 1}{x^{12}}$$ which is $378000000$ at $x=0$. Is there a simpler method to doing this by hand?

Chegg seems to agree that the answer is $378000000$

• Using Taylor series for $\cos(5x^2)$ what will you get? – Easy Apr 11 '13 at 8:32
• Isn't that $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!} (5x^2)^{2n}$? – Snakes and Coffee Apr 11 '13 at 8:36
• Sub in this expression and do the derivative, and see what happens. – Easy Apr 11 '13 at 8:40
• Do you just plug in $n=9$ or is it $n=3$? – Snakes and Coffee Apr 11 '13 at 8:42

\begin{align}f(x)=\frac{\cos{5 x^2}-1}{x^3} &= \sum_{k=1}^{\infty} \frac{(-1)^k}{(2 k)!} 5^{2 k} x^{4 k-3}\\ &= -\frac{25}{2} x + \frac{625}{24}x^5 - \frac{5^6}{6!}x^9+... \end{align}
The 9th derivative at $x=0$ is the coefficient of $x^9$ times $9!$:
$$f^{(9)}(0) = - \frac{5^6}{6!} 9! = -9 \cdot 8 \cdot 7 \cdot 5^6$$
• Why doesn't this match up with the brute force computed value of $f^{9}(0)=378000000$? – Snakes and Coffee Apr 11 '13 at 9:02
• @user10676: you have to take a limit as %x \rightarrow 0$. The terms that would blow up should cancel. – Ron Gordon Apr 11 '13 at 9:14 Use:$$\cos(5x^2)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!} (5x^2)^{2n}$$ $$\cos(5x^2)=1-\dfrac{(5x^2)^2}{2!}+\dfrac{(5x^2)^4}{4!}-\dots$$ $$\cos(5x^2)-1=-\dfrac{(5x^2)^2}{2!}+\dfrac{(5x^2)^4}{4!}-\dots$$ Now $$\dfrac{\cos(5x^2)-1}{x^3} =-\dfrac{25x^1}{2!}+\dfrac{625x^5}{4!}-\dots$$ can be easily seen to be dependent on$x$with coff($x^9$)=-$5^6/6!$. Now you can see what will you get when differentiate this 9 times. After differentiating 9 times the the term independent of x will be$-\dfrac{9!\times 5^6}{6!}\$ .